Prove that there are four distinct real number $x,y,z,w$,such $|xz+yw|\ge |\sqrt{5}(xw-yz)|$ For any nine  distinct real numbers,there exsit four  distinct real number $x,y,z,w$,such
$$|xz+yw|\ge |\sqrt{5}(xw-yz)|$$
I think can use  pigeonhole principle to solve it?Thanks 
 A: It seems that actually just $7$ real numbers suffice, though I have had to use a calculator, so some clever hand written argument may suffice for $9$ real numbers.
First lets consider when $|xy+zw|\ge|\sqrt{5}(xw-yz)|$. I apologise for the lack of diagram - you may wish to draw one as you read.
We have $xy+zw=\binom{x}{z}\cdot\binom{y}{w}=|A||B|\mathrm{cos}(\theta)$ where $\theta$ is the angle between $A=\binom{x}{z}$ and $B=\binom{y}{w}$ and $xw-yz=\binom{x}{z}\cdot\binom{w}{-y}$. Notice that $\binom{w}{-y}$ can be obtained by rotating $\binom{y}{w}$ $90^\circ$ clockwise (it turns out in this question that degrees will be easier than radians). Therefore $|\binom{x}{z}\cdot\binom{w}{-y}|=|A||B||\mathrm{cos}(90^\circ-\theta)|=|A||B||\mathrm{sin}(\theta)|$. Putting this together $|xw-yz|/|xy+zw|=|\mathrm{sin}(\theta)/\mathrm{cos}(\theta)|=|\mathrm{tan}(\theta)|$ so the question is instead when is $|\mathrm{tan}(\theta)|\le\frac{1}{\sqrt{5}}$.
I used a calculator, but there may be a clever hand written calculation you could use instead. $\mathrm{tan}^{-1}(\frac{1}{\sqrt{5}})\approx 24.09^\circ$. In particular, if $|\theta|\le 24^\circ$ or $|180^\circ-\theta|\le 24^\circ$ then the inequality holds. 
Now, as you suspected we use the pigeonhole principle.
From the $7$ real numbers, choose three disjoint pairs $(x_1,x_2),(x_3,x_4),(x_5,x_6)$ such that each number from a given pair has the same sign - it is easy to show that this is possible! In particular each $\binom{x_i}{x_{i+1}}$ is in either the upper right or lower left quadrant of the real plane - note also that $x_i\ne x_{i+1}$ so they do not lie on the line $x=y$.
Now we split the union of these quadrants (minus the line $x=y$) into $4$ sets. $$A=\left\{\binom{x}{y}|x=0\hspace{0.2cm}or\hspace{0.2cm}\mathrm{tan}^{-1}(y/x)>77.5^\circ\right\}$$ $$B=\left\{\binom{x}{y}|45^\circ<\mathrm{tan}^{-1}(y/x)\le77.5^\circ\right\}$$ $$C=\left\{\binom{x}{y}|22.5^\circ\le \mathrm{tan}^{-1}(y/x)<45^\circ\right\}$$ $$D=\left\{\binom{x}{y}|0^\circ\le\mathrm{tan}^{-1}(y/x)<22.5^\circ\right\}$$
Notice that any two points lying in one of these for sets create an angle at the origin $\theta$ with $|\theta|\le 22.5^\circ$ or $|180^\circ-\theta|\le 22.5^\circ$. Therefore, to find $4$ real numbers which satisfy the inequality it suffices to use $4$ to define two points in one of these sets.
From each pair place the point $\binom{x_i}{x_{i+1}}$ on the plane. By the pidgeonhole principle, we must have two points lie in either $A\cup D$ or $B\cup C$. If a point $\binom{x}{y}$ lies in $C$ or $D$ then $\binom{y}{x}$ lies in $B$ or $A$ respectively. This means that by swapping $x_i$ with $x_{i+1}$ if necessary we obtain two points in $A$ or two points in $B$.
